Weakening the hypotheses in the Duffin-Schaeffer conjecture? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:32:56Z http://mathoverflow.net/feeds/question/63514 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63514/weakening-the-hypotheses-in-the-duffin-schaeffer-conjecture Weakening the hypotheses in the Duffin-Schaeffer conjecture? Stanley Yao Xiao 2011-04-30T08:52:14Z 2011-11-18T22:39:49Z <p>The Duffin-Schaeffer conjecture is a conjecture in metric number theory, which asks for a given function $f : \mathbb{R} \rightarrow \mathbb{R}^+$ the measure of the set of real numbers $\alpha$ such that the inequality $$\displaystyle \left | \alpha - \frac{p}{q} \right| &lt; \frac{f(q)}{q}$$ has infinitely many solutions in integers $p,q$ with $\gcd(p,q) = 1$. The conjecture asserts that the set of solutions $\alpha$ has full measure if and only if $\displaystyle \sum_{q=1}^\infty \frac{f(q)}{q} \phi(q)$ diverges, where $\phi(q)$ is the Euler totient function. </p> <p>Let the set of solutions (which depends on the function $f$) be denoted $E_f$. Then it is known (Haynes, Pollington, Velani <a href="http://arxiv.org/abs/0811.1234" rel="nofollow">http://arxiv.org/abs/0811.1234</a>) that a sufficient condition for $m(\mathbb{R} \setminus E_f) = 0$ is for $\displaystyle \sum_{q=1}^\infty \left(\frac{f(q)}{q}\right)^{1 + \epsilon} \phi(q) = \infty$. My question concerns other possible sufficient conditions. In particular, we know (from Duffin and Schaeffer themselves) that it is not sufficient for $\displaystyle \sum_{q=1}^\infty f(q) = \infty$. Is it sufficient for the sum $\displaystyle \sum_{q=1}^\infty \frac{f(q)}{q^\epsilon}$ to diverge for any $\epsilon > 0$? What about $\displaystyle \sum_{q=1}^\infty \frac{f(q)}{\log^C(q)} = \infty$ for some $C \geq 1$?</p> http://mathoverflow.net/questions/63514/weakening-the-hypotheses-in-the-duffin-schaeffer-conjecture/81289#81289 Answer by Alan Haynes for Weakening the hypotheses in the Duffin-Schaeffer conjecture? Alan Haynes 2011-11-18T22:39:49Z 2011-11-18T22:39:49Z <p>To my knowledge here is the best known sufficient condition:</p> <p>Let $c$ be any positive constant and let $g: [0,\infty)\rightarrow [0,\infty)$ be defined by $g(0)=0$, $$g(x)=x \exp\left(-c (\log (- \log x))(\log \log (-\log x)) \right) \quad \text{ if } ~ 0 &lt; x &lt; 1,$$ and $g(x)=1$ if $x\ge 1$. If $$\sum_{q=1}^{\infty} g\left(\frac{f(q)}{q}\right) \varphi(q)=\infty$$ then $m(\mathbb{R}\setminus E_f)=0$. In particular this implies that the Duffin-Schaeffer Conjecture is true for any $f$ which satisfies $$\sum_{q=16}^{\infty} \frac{\varphi(q) f(q)}{q \exp(c(\log \log q)(\log \log \log q))}=\infty$$ for some $c>0$. Even more particularly, since $(\log \log q)(\log \log \log q)\ll \log q$ this means that if $$\sum_{q=1}^\infty\frac{f(q)}{q^\epsilon}$$ diverges for some $\epsilon >0$ then the Duffin-Schaffer Conjecture holds. However if you replace this sum with the second sum that you asked about (i.e. the one with the power of a logarithm in the denominator), then the answer is not known.</p> <p>By the way the $1+\epsilon$ result that you mentioned is actually a corollary of a theorem that was originally proved by Glyn Harman (it follows from Theorem 3.7(iii) in his book Metric Number Theory). However the result in the Haynes, Pollington, Velani paper is actually a little stronger.</p>